Question

it's a integration...plz solve it​

Answer 1

$\rm \longrightarrow \displaystyle \int \rm\frac{2x + 3}{3x - 4} \: dx$

$\rm let \: \: 3x - 4 = t \\ \rightarrow \rm 3x = t + 4 \\ \rightarrow \rm x = \frac{t + 4}{3} \\ \\ \rm Also, \: 3 \: dx = dt \\ \rm \rightarrow dx = \dfrac{dt}{3}$

$\rm \longrightarrow \displaystyle \int \rm\dfrac{2 \bigg( \dfrac{t + 4}{3} \bigg )+ 3}t \: \dfrac{dt}{3}$

$\rm \longrightarrow \dfrac{1}{3} \displaystyle \int \rm\dfrac{2 \bigg( \dfrac{t + 4}{3} \bigg )+ 3}t \: dt$

$\rm \longrightarrow \dfrac{1}{3} \displaystyle \int \rm\dfrac{ \bigg( \dfrac{2t + 8}{3} \bigg )+ 3}t \: dt$

$\rm \longrightarrow \dfrac{1}{3} \displaystyle \int \rm\dfrac{ \bigg( \dfrac{2t + 8 + 9}{3} \bigg )}t \: dt$

$\rm \longrightarrow \dfrac{1}{3} \displaystyle \int \rm\dfrac{ \bigg( \dfrac{2t + 17}{3} \bigg )}t \: dt$

$\rm \longrightarrow \dfrac{1}{3} \displaystyle \int \rm\dfrac{ \bigg( {2t + 17} \bigg )}{3t }\: dt$

$\rm \longrightarrow \dfrac{1}{9} \displaystyle \int \rm\dfrac{ {2t + 17} }{t }\: dt$

$\rm \longrightarrow \dfrac{1}{9} \displaystyle \int \rm\dfrac{ {2t } }{t }\: dt +\dfrac{1}{9} \displaystyle \int \rm \dfrac{ {17} }{t }\: dt$

$\rm \longrightarrow \dfrac{2}{9} \displaystyle \int 1 \: \rm dt +\dfrac{17}{9} \displaystyle \int \rm \dfrac{ 1 }{t }\: dt$

$\rm \longrightarrow \dfrac{2}{9} \displaystyle t + \rm \dfrac{17}{9}ln t + c$

Now, put t = 3x-4

$\rm \longrightarrow \dfrac{2}{9} (3x - 4) + \rm \dfrac{17}{9}ln (3x - 4)+ c$

Answer :-

$\rm \longrightarrow \dfrac{2}{9} (3x - 4) + \rm \dfrac{17}{9}ln (3x - 4)+ c$

or

$\rm \longrightarrow \dfrac{2x}{3} + \rm \dfrac{17}{9}ln (3x - 4)+ c$